Hi,
trying to mix lambda expresions and list comprehension
doesn't seem to work.
--- [lambda x:x+y for y in range(10)][9](2)
11 [lambda x:x+y for y in range(10)][4](2)
11
---
I expected the second expression to return 6.
What did I do wrong? Any hints?
Thanks
- Peter 4  3041 
Try this: [lambda x, y=y:x+y for y in range(10)][4](2)
6
It is important to bind y in a closure at the time
the lambda is defined. Otherwise, y remains unbound
until you invoke the function call. At that time, the most
recent value of y is the last value in the range loop (namely, 9).
Raymond Hettinger
"Peter Barth" <pe*********@ t-online.de> wrote in message
news:6f******** *************** ***@posting.goo gle.com... Hi,
trying to mix lambda expresions and list comprehension
doesn't seem to work.
--- [lambda x:x+y for y in range(10)][9](2) 11 [lambda x:x+y for y in range(10)][4](2)
11
---
I expected the second expression to return 6.
What did I do wrong? Any hints?
Thanks
- Peter
pe*********@t-online.de (Peter Barth) wrote in message news:<6f******* *************** ****@posting.go ogle.com>... Hi,
trying to mix lambda expresions and list comprehension
doesn't seem to work.
--- [lambda x:x+y for y in range(10)][9](2) 11 [lambda x:x+y for y in range(10)][4](2) 11
---
I expected the second expression to return 6.
What did I do wrong? Any hints?
Thanks
- Peter
It is a scope issue. The last value for y is used for all
the created lambdas. All lambdas users are bitten by that,
soon or later. The solution is to make y local to the
lambda function, with the optional argument trick:
[lambda x,y=y:x+y for y in range(10)][4](2)
6
Michele
Thanks a lot, works fine.
However, the solution does not really feel "pythonesqu e".
Is it considered a usability bug or fine as is?
- Peter
"Raymond Hettinger" <vz******@veriz on.net> wrote in message news:<hM******* ********@nwrdny 01.gnilink.net> ... Try this:
[lambda x, y=y:x+y for y in range(10)][4](2) 6
It is important to bind y in a closure at the time
the lambda is defined. Otherwise, y remains unbound
until you invoke the function call. At that time, the most
recent value of y is the last value in the range loop (namely, 9).
Raymond Hettinger
"Peter Barth" <pe*********@ t-online.de> wrote in message
news:6f******** *************** ***@posting.goo gle.com... Hi,
trying to mix lambda expresions and list comprehension
doesn't seem to work.
--->> [lambda x:x+y for y in range(10)][9](2)
11>> [lambda x:x+y for y in range(10)][4](2)
11
---
I expected the second expression to return 6.
What did I do wrong? Any hints?
Thanks
- Peter
bo**@oz.net (Bengt Richter) wrote in message news:<bf******* ***@216.39.172. 122>... On 15 Jul 2003 05:41:02 -0700, mi**@pitt.edu (Michele Simionato) wrote:
pe*********@ t-online.de (Peter Barth) wrote in message news:<6f******* *************** ****@posting.go ogle.com>... Hi,
trying to mix lambda expresions and list comprehension
doesn't seem to work.
---
>>> [lambda x:x+y for y in range(10)][9](2) 11 >>> [lambda x:x+y for y in range(10)][4](2)
11
---
I expected the second expression to return 6.
What did I do wrong? Any hints?
Thanks
- Peter
It is a scope issue. The last value for y is used for all
the created lambdas. All lambdas users are bitten by that,
soon or later. The solution is to make y local to the
lambda function, with the optional argument trick:
> [lambda x,y=y:x+y for y in range(10)][4](2)
6
or you could capture y as constants in the lambdas ;-)
>>> [eval('lambda x:x+%s'%y) for y in range(10)][4](2)
6
Regards,
Bengt Richter
Thanks to God, there is a smile in your post!!
;)
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